Simplify the following expression: $y = \dfrac{-5x^2- 9x+18}{-5x + 6}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-5)}{(18)} &=& -90 \\ {a} + {b} &=& &=& {-9} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-90$ and add them together. Remember, since $-90$ is negative, one of the factors must be negative. The factors that add up to ${-9}$ will be your ${a}$ and ${b}$ When ${a}$ is ${6}$ and ${b}$ is ${-15}$ $ \begin{eqnarray} {ab} &=& ({6})({-15}) &=& -90 \\ {a} + {b} &=& {6} + {-15} &=& -9 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-5}x^2 +{6}x) + ({-15}x +{18}) $ Factor out the common factors: $ x(-5x + 6) + 3(-5x + 6)$ Now factor out $(-5x + 6)$ $ (-5x + 6)(x + 3)$ The original expression can therefore be written: $ \dfrac{(-5x + 6)(x + 3)}{-5x + 6}$ We are dividing by $-5x + 6$ , so $-5x + 6 \neq 0$ Therefore, $x \neq \frac{6}{5}$ This leaves us with $x + 3; x \neq \frac{6}{5}$.